hdoj1084What Is Your Grade?（结构体+sort）

Description
“Point, point, life of student!”
This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course.
There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50.
Note, only 1 student will get the score 95 when 3 students have solved 4 problems.
I wish you all can pass the exam!
Come on!

Input
Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p.
A test case starting with a negative integer terminates the input and this test case should not to be processed.

Output
Output the scores of N students in N lines for each case, and there is a blank line after each case.

Sample Input
4
5 06:30:17
4 07:31:27
4 08:12:12
4 05:23:13
1
5 06:30:17
-1

Sample Output
100
90
90
95

100

自己的错误代码：

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;struct st{int number;char time[10];int k;int scorce;}a[101];bool cmp(st x,st y){if(x.number!=y.number)return x.number>y.number;elsereturn strcmp(x.time,y.time)<0;}bool cmo(st x,st y){return x.k<y.k;}int main(){int n,c[6],p[6],t[6]={50,60,70,80,90,100};while(scanf("%d",&n),n!=-1){memset(a,0,sizeof(a));memset(c,0,sizeof(c));memset(p,0,sizeof(p));for(int i=0;i<n;i++){scanf("%d",&a[i].number);if(a[i].number!=0)scanf("%s",a[i].time);a[i].k=i;c[a[i].number]++;a[i].scorce=t[a[i].number];}for(int i=1;i<4;i++){if(c[i]==1)p[i]=1;elsep[i]=c[i]/2;}sort(a,a+n,cmp);for(int i=0;i<n;i++){if(a[i].number<5&&a[i].number>0&&p[a[i].number]>0){a[i].scorce+=5;p[a[i].number]--;}}sort(a,a+n,cmo);for(int i=0;i<n;i++)printf("%d\n",a[i].scorce);printf("\n");}return 0;}

这个思路结果对，但是会超时。

正确代码：

#include<stdio.h>  #include<string.h>  #include<algorithm>  using namespace std;  struct Stu  {      int ans,g,id;      char ti[25];  }a[111];  bool cmp1(Stu a,Stu b)  {      if(a.ans!=b.ans)          return a.ans>b.ans;      return strcmp(a.ti,b.ti)<0;  }  bool cmp2(Stu a,Stu b)  {      return a.id<b.id;  }  int main()  {      int n,i;      while(scanf("%d",&n)&&n!=-1)      {          int s1,s2,s3,s4;          s1=s2=s3=s4=0;          for(i=0;i<n;i++)          {              scanf("%d%s",&a[i].ans,a[i].ti);              a[i].id=i;              if(a[i].ans==4) s4++;              if(a[i].ans==3) s3++;              if(a[i].ans==2) s2++;              if(a[i].ans==1) s1++;          }          sort(a,a+n,cmp1);          int p,q,r,s;          p=q=r=s=0;          for(i=0;i<n;i++)          {              if(a[i].ans==5)                  a[i].g=100;              if(a[i].ans==4)              {                  if(p<s4/2)                  {                      a[i].g=95;                      p++;                  }                  else                      a[i].g=90;              }              if(a[i].ans==3)              {                  if(q<s3/2)                  {                      a[i].g=85;                      q++;                  }                  else                      a[i].g=80;              }              if(a[i].ans==2)              {                  if(r<s2/2)                  {                      a[i].g=75;                      r++;                  }                  else                      a[i].g=70;              }if(a[i].ans==1)              {                  if(s<s1/2)                  {                      a[i].g=65;                      s++;                  }                  else                      a[i].g=60;              }              if(a[i].ans==0)                  a[i].g=50;          }          sort(a,a+n,cmp2);          for(i=0;i<n;i++)              printf("%d\n",a[i].g);          printf("\n");      }      return 0;  }  

思路：对于相同的数，记录有多少个，然后在判断是否属于前面一半，如果是输出对应值+5；主要复杂的就是判断是否属于前半部分，还要注意0道题还有输出格式。